Solve Input Math Problem: y``+y`+y=r(t)

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y`` +y`+y=r(t)
why's r(t) the input and not t?
 
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asdf1 said:
y`` +y`+y=r(t)
why's r(t) the input and not t?

Because the right hand side of that equation denotes a function of t.

r(t) could be t but it can also be every other function that depends on t.

Basically, r(t) is the most general form to denote "a function of t"

marlon
 
That's not really mathematics- it's "Engineer speak".

From the point of view of an Engineer, a differential equation is a machine to which you supply an "input" and get an "output". The differential operator y"+ y'+ y is the machine. Whatever function you have on the right hand side is the "input" (which varies with t) and y(t) satisfying the equation is the "output".

(Edited- thanks, Marlon.)
 
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HallsofIvy said:
That's not really mathematics- it's "Engineer speak".
From the point of view of an Engineer, a differential equation is a machine to which you supply an "input" and get an "output". The differential operator y"+ y' is the machine. Whatever function you have on the right hand side is the "input" (which varies with t) and y(t) satisfying the equation is the "output".

The differential operator is not y''+y' but y''+y'+y

marlon
 
thank you very much!
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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